# How does beam pinned at one end resist moment?

This beam is constrained by a pin at one end, so it is free to rotate around that point. If a moment M is applied, the beam cannot move since any rotation would involve motion of the COM, which is not possible since there are no forces.

So if the com cannot angularly accelerate when there is only a moment applied, and it cannot move without rotating since it is pinned, it must be at static equilibrium.

If you take your moments about the constraint, there is no way to get a moment to cancel out M! But if you take your moments about com and solve for a force at the constraint to cancel out the moment, then you have an unbalanced force!

How is this possible?

A rigid body with a net moment applied to it is clearly NOT in static equilibrium. Therefore it moves. The only thing that may be puzzling is how it moves, consistent with Newton's laws of motion.

To get started untangling the confusion, forget about the pin. You now have a rigid body with no constraints on its motion, and a moment applied to it. The moment will cause angular acceleration about the COM, and since there are no net forces, the COM will not move.

But in your example, this can't happen, because the end of the bar is constrained not to move by the pin. The pin will apply a constraint force to the bar, perpendicular to it, to prevent it moving.

That constraint force will cause the COM to accelerate upwards, such that the motion of the bar is a rotation about its end, not about its COM. If the bar as shown in your picture is along the $$x$$ axis, then $$\ddot y = (l/2)\ddot \theta$$ where $$l$$ is the length of the bar, $$y$$ is the acceleration of the COM in the $$y$$ direction, and $$\theta$$ is the angle of rotation of the bar.

That is where the "force to accelerate the COM" is coming from. As the bar rotates, the reaction force will change, because there will also be a component along the length of the bar, causing the centripetal acceleration that makes the COM move in a circle.

Note that the reaction forces do no work, and therefore the easiest way to formulate the equation of motion is equate the work done by the moment and the kinetic energy of the bar, and use the constraint equation given earlier to link the translational and rotational motion of the bar.

The hinge constrains the beam by applying forces to it. The only way to say it applies zero force is to remove it.

So we can assume that the hinge is able to supply a force and in this case, the force is whatever is required to accelerate the COM.

• Exactly, the pin forces are what they need to be for the allowed motion kinematically (rotation about the pin). – ja72 Dec 16 '18 at 3:10

Your free body diagram is missing the pin reactions.

When you include those, you see that the center of mass indeed moves and a pin joint will not resist moments.

The solution of the equations of motion are

\begin{aligned} F_x & = -m c \dot{\theta}^2 \\ F_y & = \frac{m c}{I_C + m c^2} M \\ \ddot{\theta} & = \frac{M}{I_C + m c^2} \end{aligned}

where $$\theta$$ and its derivatives is the swing angle, $$m$$ is the mass, $$I_C$$ is the mass moment of inertia about the COM and $$c$$ is the distance between the pin and the COM.

As you can see there is rotational acceleration $$\ddot{\theta}$$ due to the torque $$M$$. Also, this torque generates the reaction force $$F_y$$ which in turn accelerates the center of mass.